What is the sum from k=1 to infinity of k/(k^2-1) ?

The sum from k=1 to infinity of k/(k^2-1) is diverges

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sum from k=1 to infinity of k/(k^2-1)

k = 1 \sum \infty \frac{k}{k ^{2} - 1}

diverges

Solution steps

k = 2 \sum \infty \frac{k}{k ^{2} - 1}

Apply Series Comparison Test:

diverges

= diverges

n = 1 \sum \infty \frac{1 + cos ( n )}{n}

n = 0 \sum \infty 1 - e^{- n}

k = 1 \sum \infty \frac{k}{k ^{2} + 4}

n = 2 \sum \infty \frac{1}{( n - 1 ) ^{4}}

n = 0 \sum \infty 5 (- \frac{3}{2})^{n}

What is the sum from k=1 to infinity of k/(k^2-1) ?
The sum from k=1 to infinity of k/(k^2-1) is diverges